Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const int MAXN = 100001, LOG = 17;
vector<int> e[MAXN];
int up[LOG + 1][MAXN];
int h[MAXN];
vector<int> x[LOG + 1][MAXN];
void dfs(int v, int p) {
  up[0][v] = p;
  for (int u : e[v]) {
    if (u == p) continue;
    h[u] = h[v] + 1;
    dfs(u, v);
  }
}
int jmp(int v, int d) {
  for (int i = (int)(LOG)-1; i >= 0; --i) {
    if ((1 << i) <= d) {
      v = up[i][v];
      d -= 1 << i;
    }
  }
  return v;
}
int lca(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  u = jmp(u, h[u] - h[v]);
  if (u == v) return u;
  for (int i = (int)(LOG)-1; i >= 0; --i) {
    int uu = up[i][u], vv = up[i][v];
    if (uu != vv) u = uu, v = vv;
  }
  return up[0][u];
}
vector<int> merg(const vector<int> &a, const vector<int> &b, int A) {
  vector<int> v;
  int i = 0, j = 0;
  while (i < a.size() && j < b.size() && v.size() < A) {
    if (a[i] == b[j]) {
      v.push_back(a[i]);
      ++i;
      ++j;
      continue;
    }
    if (a[i] < b[j])
      v.push_back(a[i++]);
    else
      v.push_back(b[j++]);
  }
  while (i < a.size() && v.size() < A) v.push_back(a[i++]);
  while (j < b.size() && v.size() < A) v.push_back(b[j++]);
  return v;
}
void upq(int v, int d, int a, vector<vector<int> *> &res) {
  for (int i = (int)(LOG)-1; i >= 0; --i) {
    if ((1 << i) <= d) {
      res.push_back(&x[i][v]);
      v = up[i][v];
      d -= 1 << i;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int N, M, Q;
  scanf("%d%d%d", &N, &M, &Q);
  for (int i = 0; i < (int)(N - 1); ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    --x;
    --y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  for (int i = 0; i < (int)(M); ++i) {
    int v;
    scanf("%d", &v);
    --v;
    x[0][v].push_back(i);
  }
  for (int i = 0; i < (int)(N); ++i) {
    sort((x[0][i]).begin(), (x[0][i]).end());
    if (x[0][i].size() > 10) x[0][i].resize(10);
  }
  dfs(0, 0);
  for (int i = 0; i < (int)(LOG); ++i)
    for (int j = 0; j < (int)(N); ++j) {
      up[i + 1][j] = up[i][up[i][j]];
      x[i + 1][j] = merg(x[i][j], x[i][up[i][j]], 10);
    }
  cerr << "built\n";
  for (int q = 0; q < (int)(Q); ++q) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    --u;
    --v;
    int w = lca(u, v);
    vector<vector<int> *> res;
    upq(u, h[u] - h[w] + 1, a, res);
    upq(v, h[v] - h[w], a, res);
    vector<int> ans;
    vector<int> ind(res.size());
    set<pair<int, int> > s;
    for (int i = 0; i < (int)(res.size()); ++i) {
      if (!res[i]->empty()) s.insert(make_pair((*res[i])[0], i));
    }
    while (!s.empty() && ans.size() < a) {
      int x = s.begin()->first, j = s.begin()->second;
      ans.push_back(x);
      s.erase(s.begin());
      ++ind[j];
      if (ind[j] < res[j]->size()) s.insert(make_pair((*res[j])[ind[j]], j));
    }
    printf("%d", (int)ans.size());
    for (int x : ans) printf(" %d", x + 1);
    printf("\n");
  }
  return 0;
}